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\title{Homework 1: L'Hospital Law}


\author{张凌帆 ZhangLingfan \\ 数学与应用数学 3190104913}

\begin{document}

\maketitle
This is a problem from the field of mathematical analysis. In mathematical analysis, the limit of the ratio of infinity to small quantity may or may not exist. Therefore, we collectively call this limit "infinitive limit" and record it as
$\frac{0}{0}$ or $\frac{\infty}{\infty}$. Now we will use the derivative as a tool to study the limit of indefinite form. This method is collectively referred to as L'Hospital Law. Cauchy's mean value theorem is the theoretical basis for establishing L'Hospital Law.

\section{The Statement and the proof for $\frac{0}{0}$}
\subsection{$\frac{0}{0}$}
If the function $f$ and $g$ satisfy the following conditions:
\begin{itemize}
\item {\bf (1)} $\lim_{x \to x_0}f(x)=\lim_{x \to x_0}g(x)=0$;
\item {\bf (2)} In the neighborhood $U^{0}(x_0)$ of $x_0$ ,$f(x)$ and $g(x) $are both differentiable.
\item {\bf (3)} $\lim_{x\to x_0} \frac{f'(x)}{g'(x)}=A$ where A can be real numbers or complex numbers.
then,we have
$\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f'(x)}{g'(x)}=A$
\end{itemize}

\subsection{Proof}
Proof of 1.1:
Define 
\begin{equation}
f(x_0)=g(x_0)=0
\label{eq::continous}
\end{equation}
such that $f$ and $g$ are both continous on teh point $x_0$. For any point $x\in U^{0}(x_0)$, apply the Cuachy mean value thoerem in the interval $[x_0.x]$(or $[x.x_0]$), we have 
\begin{equation}
\frac{f(x)-f(x_0)}{g(x)-g(x_0)}=\frac{f(\xi)}{g(\xi)}
\label{eq::cauchy}
\end{equation}
then with (\ref{eq::continous}) and (\ref{eq::cauchy}) we can know that, 
\begin{equation}
\frac{f(x)}{g(x)}=\frac{f'(\xi)}{g'(\xi)}(\xi is in between x_0 and x)
\label{conclusion 1}
\end{equation}
when $x\to x_0$, we can know from (\ref{conclusion 1}) that,
\begin{equation}
\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_\frac{f'(\xi)}{g'(\xi)}=\lim_{x\to x_)}\frac{f'(x)}{g'(x)}=A
\label{final result}
\end{equation}
Then, (\ref{final result}) is the final result. The Theorem is proved.


\section{The Statement and the proof for $\frac{\infty}{\infty}$}

\subsection{$\frac{\infty}{\infty}$}
If the function $f$ and $g$ satisfy the following conditions:
\begin{itemize}
\item {\bf (1)} $\lim_{x \to x_0}g(x)=\infty$;
\item {\bf (2)} In the right neighborhood $U^{0}_+(x_0)$ of $x_0$ ,$f(x)$ and $g(x) $are both differentiable, and $g'(x_0)\neq 0$.
\item {\bf (3)} $\lim_{x\to x_0} \frac{f'(x)}{g'(x)}=A$ where A can be real numbers or $\infty$.
then,we have
$\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f'(x)}{g'(x)}=A$
\end{itemize}

\subsection{Proof}
Proof:
If A is a real number, we can see, for any positive number $\epsilon$,there exists $a_1 \in U^0_+(x_0)$, for any $x$ satisfying $x_0\le x\le x_1$, there is 
\begin{equation}
A-\frac{\epsilon}{2}\le \frac{f'(x)}{g'(x)}\le A-\frac{\epsilon}{2}
\label{epsilondelta for 2}
\end{equation}
then, according to the Cuachy mean value thoerem we have 
\begin{equation}
A-\frac{\epsilon}{2}\le (\frac{f(x)}{g(x)}-\frac{f(x_1)}{g(x)})(\frac{g(x)}{g(x)-g(x_1)})=\frac{f(x)-f(x_1)}{g(x)-g(x_1)}=\frac{f'(\epsilon)}{g'(\epsilon)}\le A+\frac{\epsilon}{2}
\label{using cauchy for 2}
\end{equation}
Then according to (\ref{using cauchy for 2})  and the locally sign preserving property, we can see that
\begin{equation}
(1-\frac{g(x_1)}{g(x)})(A-\frac{\epsilon}{2})+\frac{f(x_1)}{g(x)}\le \frac{f(x)}{g(x)}\le (1-\frac{g(x_1)}{g(x)})(A-\frac{\epsilon}{2})+\frac{f(x_1)}{g(x)}
\label{c for 2}
\end{equation}
Accoriding to (\ref{c for 2}) and $\lim_{x\to x_{0}^{+}}g(x)=\infty$, then
\begin{equation}
\lim_{x\to x_{0}^+}((1-\frac{g(x_1)}{g(x)})(A-\frac{\epsilon}{2})+\frac{f(x_1)}{g(x)})=A-\frac{\epsilon}{2}
\lim_{x\to x_{0}^+}((1-\frac{g(x_1)}{g(x)})(A+\frac{\epsilon}{2})+\frac{f(x_1)}{g(x)})=A+\frac{\epsilon}{2}
\label{aigu}
\end{equation}
And since there is locally sigh preserving property, when $x_0\le x\le x_0+\delta$, there is 
\begin{equation}
A-\epsilon\le \frac{f(x)}{g(x)}\le A-\epsilon
\label{yes}
\end{equation}
From (\ref{yes}) we get the final result:
\begin{equation}
\lim_{x\to x_0^+}\frac{f(x)}{g(x)}=A
\label{final 2}
\end{equation}
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